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Math Help - Combinatorics

  1. #1
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    Combinatorics

    Jack has a test with a total of 15 questions. He has to answer 12 of these questions.

    In how many ways can he choose which questions to answer if he at least has to choose 3 out of the 5 first questions?

    Thank you for taking the time to read my post.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MatteNoob View Post
    Jack has a test with a total of 15 questions. He has to answer 12 of these questions.

    In how many ways can he choose which questions to answer if he at least has to choose 3 out of the 5 first questions?

    Thank you for taking the time to read my post.
    I think I've got this, so you would need to evaluate _5C_3

    _nC_r=\frac{n!}{r!(n-r)!}

    Thus, _5C_3=\frac{5!}{3!2!}=\frac{120}{6\times 2}=\color{red}\boxed{10}

    Does this make sense?

    --Chris
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    Quote Originally Posted by MatteNoob View Post
    Jack has a test with a total of 15 questions. He has to answer 12 of these questions.

    In how many ways can he choose which questions to answer if he at least has to choose 3 out of the 5 first questions?

    Thank you for taking the time to read my post.
    He must answer 3, 4, or 5 out of the first 5 questions.

    If he answers 3 out of the first 5 then he must answer 9 out of the last 10; if he answers 4 out of the first 5 then he must answer 8 out of the last 10; and if he answers 5 out of the first 5 then he must answer 7 out of the last 10.

    So the total number of possibilities is

    \binom{5}{3} \binom{10}{9} + \binom{5}{4} \binom{10}{8} + \binom{5}{5} \binom{10}{7}.
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  4. #4
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    Thank you both for taking the time to respond to my thread.

    I also thought that if he has to choose atleast 3 out of the first five questions he had to do as you said, but that is counter intuitive to me, let me explain why:

    \underbrace{\binom{5}{3} \binom{10}{9}}_{a} + \underbrace{\binom{5}{4} \binom{10}{8}}_{b} + \underbrace{\binom{5}{5} \binom{10}{7}}_{c}.

    Sorry, at further analysis I come to understand that neither a, b or c hold any identical combination of which questions he chose to answer. I appreciate your help.
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    Quote Originally Posted by MatteNoob View Post
    Thank you both for taking the time to respond to my thread.

    I also thought that if he has to choose atleast 3 out of the first five questions he had to do as you said, but that is counter intuitive to me, let me explain why:

    \underbrace{\binom{5}{3} \binom{10}{9}}_{a} + \underbrace{\binom{5}{4} \binom{10}{8}}_{b} + \underbrace{\binom{5}{5} \binom{10}{7}}_{c}.

    Sorry, at further analysis I come to understand that neither a, b or c hold any identical combination of which questions he chose to answer. I appreciate your help.
    MatteNoob,

    I am sorry, but I do not understand your question. Why is the answer counter intuitive to you?
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